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CBSE Class 12 Math 2025 All Sets Solved Paper

Question : 9 of 20

Marks: +1, -0

{a}↖{→}+{b}↖{→}+{c}↖{→}={0}↖{→}, | {a}↖{→} | ={√{37}}, | {b}↖{→} | =3

and

| {c}↖{→} | =4,

then angle between

{b}↖{→}

and

{c}↖{→}

Solution:

Given, \ {a}↖{→}+{b}↖{→}+{c}↖{→}={0}↖{→}, | {a}↖{→} | ={√{37}}, | {b}↖{→} | =3, | {c}↖{→} | =4

{b}↖{→}+{c}↖{→}=-{a}↖{→}

Squaring both sides,

| {b}↖{→} | ^{2}+ | {c}↖{→} | ^{2}+2 {b}↖{→} ⋅ {c}↖{→}= | -a | ^{2}

⇒ | {b}↖{→} | ^{2}+ | {c}↖{→} | ^{2}+2 | {b}↖{→}||{c}↖{→} | \cos θ= | {a}↖{→} | ^{2}

⇒ 9+16+2(3)(4) \cos θ=37

⇒ 24 \cos θ=37-25

⇒ 24 \cos θ=12

⇒ \;\; \cos θ=\;{1}/{2}

⇒ θ=\;{π}/{3}

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